A man is standing on a cliff and throws a ball outward 50m/s at an angle of 53 degrees. It hits the base of the cliff in 10 seconds.

Find:
a) The height of the cliff
b) the angle at which the ball goes into the sea
c) if the ball will hit a boat 150m out to sea.

To solve this problem, we can use the equations of motion for projectile motion. Projectile motion involves the vertical and horizontal motion of an object that is launched into the air at an angle.

Let's break down the problem step by step:

a) Finding the height of the cliff:
To find the height of the cliff, we need to determine the time it takes for the ball to reach the base of the cliff. Since we know the initial velocity of the ball (50 m/s), we can use the equation of vertical motion to find the time (t):
s = ut + (1/2)gt^2

In this case, the ball is thrown upwards, so the initial vertical velocity (u) is (50 m/s) * sin(53°), and the acceleration due to gravity (g) is -9.8 m/s^2 (negative because it acts downward).

Using the vertical motion equation, we can calculate t:
0 = (50 m/s) * sin(53°) * t + (1/2) * (-9.8 m/s^2) * t^2

Simplifying the equation and solving for t, we get:
(4.9 m/s^2) * t^2 - (25 m/s) * sin(53°) * t = 0

Using the quadratic formula:
t = [-(25 m/s) * sin(53°) ± sqrt((-(25 m/s) * sin(53°))^2 - 4 * (4.9 m/s^2) * 0)] / (2 * (4.9 m/s^2))

Calculating the value of t, we get two solutions: t = 0s (implying launch) and t = 8.18s.

The time taken to hit the base of the cliff is 8.18 seconds. Now, we can find the height (h) of the cliff using the same vertical motion equation:

h = (50 m/s) * sin(53°) * t + (1/2) * (-9.8 m/s^2) * t^2

Plugging in the values:
h = (50 m/s) * sin(53°) * 8.18s + (1/2) * (-9.8 m/s^2) * (8.18s)^2

Solving for h, we find that the height of the cliff is approximately 197.38 meters.

b) Finding the angle at which the ball goes into the sea:
To find the angle at which the ball goes into the sea, we need to determine the horizontal distance traveled by the ball. The horizontal motion is constant, so we can calculate it using the formula:
s = ut + (1/2)gt^2

In this case, the initial horizontal velocity (ux) is (50 m/s) * cos(53°), and the time of flight (t) is 10 seconds.

Using the horizontal motion equation, we get:
s = (50 m/s) * cos(53°) * 10s

simplifying the equation, we find:
s = 425.75 meters

So, the ball travels 425.75 meters horizontally. Now, we can find the angle (θ) at which the ball goes into the sea using the equation:
θ = arctan(horiz/vert)

θ = arctan(425.75m/197.38m)

Solving for θ, we find that the angle at which the ball goes into the sea is approximately 63.99 degrees.

c) Determining if the ball will hit a boat 150m out to sea:
To decide if the ball will hit a boat 150m out to sea, we compare the horizontal distance traveled by the ball (425.75m) with the distance to the boat (150m).

Since the horizontal distance traveled by the ball is greater than the distance to the boat, the ball will indeed hit the boat.